A G D D For, if it be possible, let E be their centre: join EC, and draw any straight line EFG meeting the circles in F and G: and because E is the centre of the circle ABC, CE is equal to EF: Again, because E is the centre of the circle CDG, CE is equal to EG: but CE was shown to be equal to EF, therefore EF is equal to EG, the less to the greater, which is impossible: therefore E is not the centre of the circles ABC, CDG. PROP. VI. THEOR. If two circles touch one another internally, they cannot have the same centre. Let the two circles ABC, CDE, touch one another internally in the point C; they have not the same centre. For, if they have, let it be F; join FC, and draw any straight line FEB meeting the circles in E and B; and because F is the centre of the circle ABC, CF is equal to FB; also, because F is the centre of the circle CDE, CF is equal to FE: but CF was shown to be equal to FB; therefore FE is equal to FB, the less to the greater, which is impossible; Wherefore F is not the centre of the circles ABC, CDE. C F E B D PROP. VII. THEOR. If any point be taken in the diameter of a circle which is not the centre, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of any others, that which is nearer to the line passing through the centre is always greater than one more remote from it; And from the same point there can be drawn only two straight lines that are equal to one another, one upon each side of the shortest line. Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre: let the centre be E; of all the straight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA is the greatest; and FD, the other part of the diameter AD, is the least; and of the others, FB is greater than FC, and FC than FG. Join BE, CE, GE; and because two sides of a triangle are greater (13. 1.) than the third, BE, EF are greater than BF; but AE is equal to EB; therefore AE and EF, that is, AF, is greater than BF: again, because C BE is equal to CE, and FE common to the triangles BEF, CEF, the two sides CE, EF are equal to the two CE, EF: but the angle BEF is greater than the angle CEF; therefore the base BF is greater (15. 1.) than the base FC; for the same reason, CF is greater than GF. Again, because GF, FE are greater (13. 1.) than EG, and EG is equal to ED; GF, FE are greater than ED; take away the common part FE, and the remainder GF is greater than the remainder FD: therefore FA is the greatest, and FD the least of all the straight lines from F to the circumference; and BF is greater than CF, and CF than GF. K H Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD: at the point F in the straight line EF, make (Prob. 9. 1.) the angle FEH equal to the angle GEF, and join FH: Then, because GE is equal to EH, and EF common to the two triangles GEF, HEF; the two sides GE, EF are equal to the two HE, EF; and the angle GEF is equal to the angle HEF; therefore the base FG is equal (4. 1.) to the base FH; but besides FH, no straight line can be drawn from F to the circumference equal to FG: for, if there can, let it be FK; and because FK is equal to FG, and FG to FH, FK is equal to FH; that is, a line nearer to that which passes through the centre, is equal to one more remote, which is impossible. COR. 1. Therefore, from any point not the centre, more than two equal lines cannot be drawn to the circumference. COR. 2. Consequently, that point from which three equal lines can be drawn is the centre. PROP. VIII. THEOR. If any point be taken without a circle, and straight lines be drawn from it to the circumference, whereof one passes through the centre; of those which fall upon the concave circumference, the greatest is that which passes through the centre; and of the rest that which is nearer to that through the centre is always greater than the more remote; But of those which fall upon the convex circumference, the least is that between the point without the circle, and the diameter; and of the rest, that which is nearer to the least is always less than the more remote: And only two equal straight lines can be drawn from the point unto the circumference, one upon each side of the least. Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC be drawn to the circumference, whereof DA passes through the centre. Of those which fall upon the concave part of the circumference AEFC, the greatest is AD, which passes through the centre; and the line nearer to AD is always greater than the more remote, viz. DE than DF, and DF than DC; but of those which fall upon the convex circumference HLKG, the least is DG, between the point D and the diameter AG; and the nearer to it is always less than the more remote, viz. DK than DL, and DL than DH. Take M the centre of the circle ABC, and join ME, MF, MC, MK, ML, MH: And because AM is equal to ME, if MD be added to each, AD is equal to EM and MD; but EM and MD are greater (13. 1.) than ED; therefore also AD is greater than ED. Again, because ME is equal to MF, and MD common to the triangles EMD, FMD: EM, MD are equal to FM, MD; but the angle EMD is greater than the angle FMD; therefore the base ED is greater (15. 1.) than the base FD. In like manner it may be shewn that FD is greater than CD. Therefore DA is the greatest; and DE greater than DF, and DF than DC. And because MK, KD are greater (13. 1.) than MD, and MK is equal to MG, the remainder KD is greater (5. Ax.) than the remainder GD, that is, GD is less than KD: And because MK, DK are drawn to the point K within the triangle MLD from M, D, the extremities of its side MD; MK, KD are less (14. 1.) than ML, LD, whereof MK is equal to ML; therefore the remainder DK is less than the remainder DL: In KGB N H M F E A like manner, it may be shewn that DL is less than DH: Therefore DG is the least, and DK less than DL, and DL than DH. Also there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least: at the point M, in the straight line MD, make the angle DMB equal to the angle DMK, and join DB; and because in the triangles KMD, BMD, the side KM is equal to the side BM, and MD common to both, and also the angle KMD equal to the angle BMD, the base DK is equal (1. 1.) to the base DB. sides DB, no straight line can be drawn from D to the circumference, equal But, beto DK for, if there can, let it be DN; then, because DN is equal to DK, and DK equal to DB, DB is equal to DN; that is, the line nearer to DG, the least, equal to the more remote, which has been shewn to be impossible. PROP. IX. THEOR. Through three given points which are not in the same straight line, one circumference of a circle may be made to pass, and but one. Let A, B, C, be three points not in the same straight line: they shall all lie in the same circumference of a circle. For, let the distances AB, BC be bisected by the perpendiculars DF, EF, which must meet in some point F; for if they were parallel, the lines DB, CB, perpendicular to them would also be parallel (Cor. 1. Th. 21. 1.), or else form but one straight line: but they meet in B, and ABC is not a straight line by hypothesis. Let then, FA, FB, and FC be drawn; then, because FA, FB meet AB at equal distances from the perpendicular, they are equal. For similar reasons FB, FC, are equal; hence the points A, B, C, are all equally distant from the point F, and consequently lie in the circumference of the circle, whose centre is F, and radius FA. It is obvious, that besides this, no other circumference can pass through the same points; for the centre, lying in the perpendicular DF bisecting the chord AB, and at the same time in the perpendicular EF bisecting the chord BC (Cor. 1. Th. 3. 3.), must be at the intersection of these perpendiculars; so that, as there is but one centre, there can be but one circumference. COR. As two circumferences cannot have three points in common, it follows that one circumference cannot cut another in more points than two. PROP. X. THEOR. If two circles cut each other, the line which passes through their centres will be perpendicular to the chord which joins the points of intersection, and will divide it into two equal parts. Let CD be the line which passes through the centres of two circles cutting each other, it will be perpendicular to the chord AB, and will divide it into two equal parts. For the line AB, which joins the points of intersection, is a chord com mon to the two circles. And if a perpendicular be erected from the middle of this chord, it will pass (Cor. 1. Th. 3. 3.) through each of the two centres C and D. But no more than one straight line can be drawn through two points; hence, the straight line which passes through the centres will bisect the chord at right angles. COR. Hence, the line joining the intersections of the circumferences of two circles, will be perpendicular to the line which joins their centres. SCHOLIUM. 1. If two circles cut each other, the distance between their centres will be less than the sum of their radii, and the greater radius will be also less than the sum of the smaller and the distance between the centres. For, CD is less (13. 1.) than CA+AD, and for the same reason, ADAC+ CD. 2. And, conversely, if the distance between the centres of two circles be less than the sum of their radii, the greater radius being at the same time less than the sum of the smaller and the distance between the centres, the two circles will cut each other. For, to make an intersection possible, the triangle CAD must be possible. Hence, not only must we have CD<AC+AD, but also the greater radius AD<AC+CD; And whenever the triangle CAD can be constructed, it is plain that the circles described from the centres C and D, will cut each other in A and B. COR. 1. Hence, if the distance between the centres of two circles be greater than the sum of their radii, the two circles will not intersect each other. COR. 2. Hence, also, if the distance between the centres be less than the difference of the radii, the two circles will not cut each other. For, AC+CD>AD; therefore, CD>AD—AC; that is, any side of a triangle exceeds the difference between the other two. Hence, the triangle is impossible when the distance between the centres is less than the difference of the radii; and consequently the two circles cannot cut each other. PROP. XI. THEOR. If two circles touch each other internally, the straight line which joins their centres being produced, will pass through the point of contact. Let the two circles ABC, ADE, touch each other internally in the point A, and let F be the centre of the circle ABC, and G the centre of the circle ADE; the straight line which joins the centres F, G, being produced, passes through the point A. H D G F E B For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG: And because AG, GF are greater (13. 1.) than FA, that is, than FH, for FA is equal to FH, being radii of the same circle; take away the common part FG, and the remainder AG is greater than the remainder GH. But AG is equal to GD, therefore GD is greater than GH; and it is also less, which is impossible. Therefore the straight line which joins the points F and G cannot fall otherwise than on the point A ; that is, it must pass through A. ། d COR. 1. If two circles touch each other internally, the distance between their centre must be equal to the difference of their radii: for the circumferences pass through the same point in the line joining the centres. |